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Current Question (ID: 20245)

Question:

$\text{In the circuit shown in the figure, the internal resistances of } 5 \text{ V and } 8 \text{ V cells are } 3 \, \Omega \text{ and } 4 \, \Omega, \text{ respectively. The current (in amperes) flows through the } 4 \, \Omega \text{ external resistor connected across points } A \text{ and } B \text{ is } \frac{1}{n} \text{ A.}$

Options:

1. $5$
2. $15$
3. $10$
4. $7$

Solution:

$\text{Hint: } E_{eq} = \frac{E_1}{r_1} - \frac{E_2}{r_2} \Bigg/ \frac{1}{r_1} + \frac{1}{r_2}$ $\text{Step 1: Find the effective EMF of the given circuit.}$ $\text{The equivalent resistance is given by:}$ $r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{3 \times 4}{3 + 4} = \frac{12}{7} \, \Omega$ $\text{The EMF of the circuit is given by:}$ $E_{eq} = \frac{E_1}{r_1} - \frac{E_2}{r_2} \Bigg/ \frac{1}{r_1} + \frac{1}{r_2}$ $\Rightarrow E_{eq} = \left( \frac{8}{4} - \frac{5}{3} \right) \times \frac{12}{7} \, \text{V} = \frac{4}{7} \, \text{V}$ $\text{Step 2: Find the current through the given resistance.}$ $\text{The equivalent diagram of the circuit is shown in the figure below:}$ $\text{Current in the external } 4 \, \Omega \text{ resistor is:}$ $I = \frac{E_{eq}}{r_{eq} + 4} = \frac{\frac{4}{7}}{\left( \frac{12}{7} \right) + 4}$ $\Rightarrow I = \frac{1}{10} \, \text{A} = \frac{1}{n} \, \text{A}$ $\Rightarrow n = 10$ $\text{Hence, option (3) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}